Answer to Question #114120 in Analytic Geometry for Randal Rodriguez

Question

Given the vertex and focus of a parabola, (4,2) and (4,5). Determine the following
a. General equation
b. Standard equation
c. Location of the opening
d. Location of the directrix (coordinate)

Answer 1

Given the vertex and focus of a parabola, (4,2) and (4,5).

Since the x-coordinates of the vertex and focus are the same, so this is a regular vertical parabola, where the x part is squared. Since the vertex is below the focus, this is a right-side up parabola and p is positive. Since the vertex and focus are 5 – 2 = 3 units apart, then p = 3.

So the equation of parabola, we get

$(x-h)^2 = 4p(y-k)$.

So, $(x-4)^2 = 12(y-2)$.

Directrix equation is y = y-coordinate of vertex - p

So Directrix equation is $y = 2 - 3 = -1$.

a) We found given parabola is

$(x-4)^2=12(y-2)$

So, General equation of parabola is: $x^2 - 8x - 12y + 40 = 0$ .

b) Standard equation of parabola is

$(x-4)^2=12(y-2)$.

c) Since the vertex is below the focus, this is a right-side up parabola.

d) Location of the Directrix: $y = -1.$